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Score: $81.34 / 100 \quad 16 / 20$ answered
Question 20
Jasmine wants to estimate the average number of patients admitted to the ICU daily across a large hospital network. Jasmine takes a random sample of 33 days and finds that the sample average is 72.8 patients. Based on previous evidence, she believes the population standard deviation is approximately 2.5 patients.
a. What is the appropriate distribution to use to construct a confidence interval based on this research? Select an answer $\checkmark$
b. Use Jasmine's findings to construct a $99 \%$ confidence interval for the appropriate population parameter. If necessary, round results to 1 decimal places.
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The 99% confidence interval for the average number of patients admitted to the ICU daily across a large hospital network is approximately \(\boxed{[71.7, 73.9]}\) patients.

Steps

Step 1 :The appropriate distribution to use in this case is the normal distribution, because the sample size is large enough (n > 30).

Step 2 :The formula for a confidence interval is: \(\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}\) where: \(\bar{x}\) is the sample mean, Z is the Z-score, which depends on the confidence level. For a 99% confidence level, the Z-score is approximately 2.576, \(\sigma\) is the population standard deviation, and n is the sample size.

Step 3 :We can plug in the given values into this formula to find the confidence interval. The sample mean is 72.8, the population standard deviation is 2.5, the sample size is 33, and the Z-score for a 99% confidence level is 2.576.

Step 4 :Calculate the confidence interval: \(72.8 \pm 2.576 \frac{2.5}{\sqrt{33}}\)

Step 5 :Calculate the lower bound: \(72.8 - 2.576 \frac{2.5}{\sqrt{33}} = 71.7\)

Step 6 :Calculate the upper bound: \(72.8 + 2.576 \frac{2.5}{\sqrt{33}} = 73.9\)

Step 7 :The 99% confidence interval for the average number of patients admitted to the ICU daily across a large hospital network is approximately \(\boxed{[71.7, 73.9]}\) patients.

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